lefteris_kaliamboswikiaorg-20200214-history
STRUCTURE OF Cd-106, Cd-108, Cd-110, Cd-111, Cd-112, Cd-113, Cd-114, Cd-116
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications Nuclear structure of cadmium (Cd) with 24 blank positions Naturally occurring cadmium (Cd) is composed of 8 isotopes. For two of them, natural radioactivity was observed, and three others are predicted to be radioactive but their decays were never observed, due to extremely long half-life times. The two natural radioactive isotopes are 113Cd (beta decay, half-life is 7.7 × 1015 years) and 116Cd (two-neutrino double beta decay, half-life is 2.9 × 1019 years). The other three are 106Cd, 108Cd (double electron capture), and 114Cd (double beta decay); only lower limits on their half-life times have been set. At least three isotopes - 110Cd, 111Cd, and 112Cd - are absolutely stable (except, theoretically, to spontaneous fission). Among the isotopes absent in the natural cadmium, the most long-lived are 109Cd with a half-life of 462.6 days, and 115Cd with a half-life of 53.46 hours. Comparing Cd of 48 protons (even number ) with Ag of 47 protons (odd number) one sees that the structure of Cd is of high symmetry, because in Cd the p48n48 makes the symmetrical alpha particle, while in Ag the alpha particle is due to the p40n40 of the square which is moved to make the symmetrical alpha particle with the p46n46. (See my STRUCTURE OF Ag-105...Ag-110 ). In the following diagram of Cd you can see the first alpha particle with p45, n45, p47 and n47 and the symmetrical one with the p46, n46, p48 and n48. You can see also the p45, n47, n46 and p48 by using the top view of the third horizontal plane. Note that the two alpha particles contribute to the creation of 8 blank positions able to receive 4(n) +4n. Especially 4(n) give the combinations of p45 and p41, of p47 and p42, of p46 and p43, and of p48 and p44. Whereas the combinations of p45 and p23, of p47 and p29, of p46 and p30, and of p48 and p24 give the extra 4n which are shown in the diagram near p23, p29, p30 and p24 respectively. In this new arrangement the squares with S = +2 -2 = 0 give the same 8 blank positions as those of Pd for receiving 8n of strong bonds. (See my STRUCTURE OF Pd-102...Pd-110 ). As in the case of Pd here the first and the sixth horizontal plane form 4 blank positions for receiving 4(n) of weak bonds, while in the second and the fifth horizontal planes can exist four blank positions for receiving 4{n} with three bonds per neutron. In other words the number N = 24 of blank positions able to receive extra neutrons is given by The two alpha particles with 4{n) + 4n The horizontal squares with 8n The first and sixth plane with 4(n) The second and fifth plane with 4{n} That is N = 4(n) +4n + 8n + 4(n) + 4{n} of opposite spins Or N = 4{n} + 12n + 8(n) = 24 extra neutrons of opposite spins STRUCTURE OF Cd-106, Cd-108, Cd -110, Cd-110, Cd-112, Cd-114 AND Cd-116 WITH S = 0 Since the two symmetrical alpha particles give S=0 and the two symmetrical squares give also S=0 one concludes that the spin S = 0 of the above stable nuclides with S= 0 is due to the extra neutrons of opposite spins . For example the Cd-106 of S=0 has 10 extra neutrons like the 4{n} and the 6of opposite spins. The Cd-108 of S=0 has 12 extra neutrons like the 4{n} and the 8[n of opposite spins. The Cd-110 of S=0 has 14 extra neutrons like the 4{n} and the 10n with opposite spins. The Cd-112 of S=0 has 16 extra neutrons like the 4{n} and the 12n with opposite spins. The Cd-114 of S=0 has 18 extra neutrons like the 4{n} the 12n and the 2(n) of opposite spins. Finally the Cd-116 of S=0 has 20 extra neutrons like the 4{n} the 12n and the 4(n) of opposite spins. Here you see that the next nuclides cannot make any stable structure because the next 4(n) of weak np bonds and the single np bonds cannot overcome the nn repulsions of short range. STRUCTURE OF Cd-111 AND Cd-113 OF S = +1/2 Since the Cd-111of S=+1/2 has 15 extra neutrons one concludes that it has 8 extra neutrons of positive spins and 7 extra neutrons of negative spins. Especially it has 2{n} + 6n of positive spins and 2{n} + 5n of negative spins. In the same way since the Cd-113 of S= +1/2 has 17 extra neutrons, one concludes that it has 9 extra neutrons of positive spins and 8 extra neutrons of negative spins. Especially it has 2{n}+ 6n + 1(n) of positive spins and 2{n} + 6n of negative spins. ' ' DIAGRAM OF Cd-96 WITH S = 0 FORMING 24 BLANK POSITIONS Here you see the additional p47n47 and also the p48n48, which make two symmetrical alpha particles of opposite spins . Whereas the p41, n41, p42, n42, p43, n43 p44, and n44 which form the central parallelepiped of opposite spins are not shown. Also 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. Note that the 4 extra neutrons (n) of the first and the sixth plane along with the 4 extra neutrons n near the p37, p38 , p39 and p40 are not shown. You can see only the 4 extra neutrons n existing under the p21 and p22 and over the p31 and p32 ' ' n40.........p40......n ' ' n........p38..........n38 H. square with n ' ' n31………p12.........n12.......p32 ' p31........n11.........p11…… n32 Sixth H. plane' ' n........ p29.........n10.........p10…… n30' ' n29………..p9..........n9 …….p30..........n Fifth H. plane' ' p47.......n27.........p8..........n8.........p28......... n48' ' n45.......p27.........n7..........p7........n28..........p46 Fourth H. plane' ' n47......p25.........n6.........p6..........n26...........p48' ' p45......n25……….p5..........n5……….p26.........n46 Third H. plane' ' n23………p4........n4………….p24......n' ' n........p23……..n3………p3………..n24 Second H.plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22 First H. plane' ' n' ........p37.. ....n37 ' ' n39.......p39........n ' H. square with n' ' ' ' ' ' TOP VIEW OF THE SQUARE OF p37n37 AND n39p39 ' THE TWO EXTRA NEUTRONS n WITH STRONG VERTICAL BONDS UNDER THE p21 AND p22 ARE ALSO SHOWN IN THE DIAGRAM , WHILE THE TWO EXTRA NEUTRONS nWITH STRONG VERTICAL BONDS UNDER THE p34 AND p33 ARE NOT SHOWN IN THE DIAGRAM ' n' ' n........P37.........n37 ' ' n39........p39.........n ' ' n' TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' HERE THE FIRST EXTRA NEUTRON (n ) MAKES THE TWO RADIAL BONDS WITH p22 AND p33 WHILE THE SECOND ONE MAKES THE TWO RADIAL BONDS WITH p21 AND p34 ' ' ' ' ' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22 ' ' n33.......p33..... (n)' TOP VIEW OF THE THIRD HORIZONTAL PLANE OF POSITIVE SPINS WITH THE 4 NUCLEONS LIKE p41, n43, n42 AND p44 WHICH MAKE THE SQUARE OF THE CENTRAL PARALLELEPIPED. HERE THE p45 AND n47 ALONG WITH THE n46 AND p40 MAKE THE SYMMETRICAL SQUARES OF THE TWO ALPHA PARTICLES. AT THE SAME PLANE ARE SHOWN ALSO THE DEUTERONS LIKE n15p15 AND p16n16. ' ' n42........p16......n16......p44 ' n47........p25........n6........p6........n26.........p48' ' p45........n25........p5........n5........p26........ n48' ' p41.......n15.......p15.......n43' Category:Fundamental physics concepts